7. Find the coordinates of the hole of \( f(x)=\frac{(x+4)(x-2)}{(x-2)(x-1)} \)

To find the coordinates of the hole for the function: \[ f(x) = \frac{(x+4)(x-2)}{(x-2)(x-1)} \] **Step 1: Simplify the Function** Notice that the factor \((x-2)\) appears in both the numerator and the denominator. We can simplify the function by canceling this common factor: \[ f(x) = \frac{(x+4)\cancel{(x-2)}}{\cancel{(x-2)}(x-1)} = \frac{x + 4}{x - 1} \quad \text{(for } x \neq 2 \text{)} \] **Step 2: Identify the Hole** A hole occurs at the \(x\)-value that makes the canceled factor zero. In this case, the canceled factor is \((x - 2)\), so: \[ x = 2 \] **Step 3: Find the \(y\)-Coordinate of the Hole** To find the corresponding \(y\)-value, substitute \(x = 2\) into the simplified function: \[ f(2) = \frac{2 + 4}{2 - 1} = \frac{6}{1} = 6 \] **Conclusion** The hole of the function is located at the point \((2, 6)\). **Answer:** After simplifying, the hole is at the point with coordinates ( 2, 6 ).